A hydrogen molecule in excited state travelling in the x-direction with kinetic energy of 1.0 eV dissociate into two hydrogen atoms. If one of themn moves perpendicular to the x-direction with a kinetic energy 1 eV. Then the energy released in te dissociation reaction will be

Options

A. 2.6 eV
B. 2 eV
C. 3.6 eV
D. 3 eV

Answer

D

Explanation

`vec(2m)^(V)`
Before collision
`p i=2mv=sqrt(4mk.E)=sqrt(4m) units`
`p_(1)y=p_(2)y=sqrt(2xxmxx1)=sqrt(2m)units`
`=[(p_(2)^(2))/(2m)+(p_(1y^(2)))/(2m)-1]ev`
energy of dissociation =`[((sqrt(4m))^(2)+(sqrt(2m))^(2))/(2m)+((sqrt(2m))^(2))/(2m)-1]ev`
`=3ev`